There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-1 are friends, Whose ID is i and i+1 are friends. These n men stand in line. Now we select an interval of men to make some group. K men in a group can create K*K value. The value of an interval is sum of these value of groups. The people of same group’s id must be continuous. Now we chose an interval of men and want to know there should be how many groups so the value of interval is max.
Input
First line is T indicate the case number.
For each case first line is n, m(1<=n ,m<=100000) indicate there are n men and m query.
Then a line have n number indicate the ID of men from left to right.
Next m line each line has two number L,R(1<=L<=R<=n),mean we want to know the answer of [L,R].
Output
For every query output a number indicate there should be how many group so that the sum of value is max.
Sample Input
1
5 2
3 1 2 5 4
1 5
2 4
Sample Output
1
2
題目：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define met(s,k) memset(s,k,sizeof s)
#define scan(a) scanf("%d",&a)
 

#define scanl(a) scanf("%lld",&a)
#define scann(a,b) scanf("%d%d",&a,&b)
#define scannl(a,b) scanf("%lld%lld",&a,&b)
#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define prin(a) printf("%d\n",a)
#define prinl(a) printf("%lld\n",a)
using namespace std;
typedef long long ll;
const  int maxn=1e5+10;
int n,m,cont,ans,ans1[maxn];
struct que
{
    int l,r,pos;
}q[maxn];
bool cmp(que a,que b)
{
    int x=a.l/cont;
    int y=b.l/cont;
    if(x!=y)return x<y;
    else return a.r<b.r;
}
int a[maxn],vis[maxn];
void removeposx(int pos)//暴力移動指標及維護答案
{
    if(vis[a[pos]-1]&&vis[a[pos]+1])ans+=1;
    else if(vis[a[pos]-1]==0&&vis[a[pos]+1]==0)ans-=1;
    vis[a[pos]]=0;
}
void addposx(int pos)
{
    if(vis[a[pos]-1]&&vis[a[pos]+1])ans-=1;
    else if(vis[a[pos]-1]==0&&vis[a[pos]+1]==0)ans+=1;
    vis[a[pos]]=1;
}
void modui(int m)
{
    int currenl=1,currenr=0;
    for (int i = 0; i <m; ++i)//注意自增和自減號的位置
    {
        while (currenr<q[i].r)addposx(++currenr);
        while (currenr>q[i].r)removeposx(currenr--);
        while (currenl<q[i].l)removeposx(currenl++);
        while (currenl>q[i].l)addposx(--currenl);
        ans1[q[i].pos]=ans;
    }
}
int main()
{
    int t;
    scan(t);
    while (t--)
    {
        scann(n,m);
        ans=0;
        met(vis,0);
        cont=(int)sqrt((double)n);
        for(int i=1;i<=n;i++)scan(a[i]);
        for(int i=0;i< m;i++)scann(q[i].l,q[i].r),q[i].pos=i;
        sort(q,q+m,cmp);
        modui(m);
        for (int j = 0; j <m; ++j)prin(ans1[j]);
    }
    return  0;
}